Kirchoff's Laws to find all currents flowing in the cct

[boo_lufc]

1) Using Kirchoff's Laws, find all the currents flowing in the cct.

2) Hence, determine the voltages across all resistors and check that each loop complies with KVL

 

[gomunkul51]

show us what you've done and where you have problems.

 

[boo_lufc]

I have not seen a problem like this before but my basic knowledge of Kirchoff would lead me to take an equivalent R for the 1k and 4k resistor to make two loops. I would then have I1 coming from the 12V source, then at the node have I3 coming down the branch toward the 10V source and I2 carrying on towards the 6V source.
I would then obtain equations for the loops as well as I1 = I2 + I3 and try and manipulate these to get values.

Would this be the right method?

 

[gomunkul51]

You need to write 2 KCL equations and 3 KVL equations in this circuit to solve it.
You will get 5 unknown currents (through 5 sections of the circuit) in 5 linear equations.

**You can't take the equivalent R here.
Basically you can take the equivalent of a pair of elements only if you know that they have either the same voltage or the same current going through them, this is not the case in this circuit.

Does it make any sense?

 

[boo_lufc]

i think so but i am unsure about directions because i have not seen 3 voltage sources used before: Here's what i came up with if you could suggest if this is correct:

KCL at node 1: -I1 -I2 + I3 = 0
KCL at node 2: -I3 -I4 + I5 = 0

KVL (left) : 12 - (I1*R1) + (I2*R2) - 10 = 0
KVL (middle): 10 - (I2*R2) - I3*R3) + (I4*R4) - 6 = 0
KVL (right): 6 - (I4*R4) + (I5*R5) = 0

Please see image for resistor and current labels and directions

 

[boo_lufc]

just hoping this cct looks better

 

[gneill]

You only need to write three KVL equations covering the three loops.

Be sure to "walk" all the way around each loop and include all currents flowing in each component.

 

[boo_lufc]

ok thanks but where have i gone wrong in the three eqns i posted?

 

[boo_lufc]

Would i rewrite as :
Loop 1: 12 - I1R1 + (I1 - I2)R2 - 10 = 0

Loop2: 10 - (I1 - I2)R2 - I2R3 + (I2 - I3)R4 - 6 = 0

Loop 3 : 6 - (I2- I3) + I3R5 = 0 ?

 

[gneill]

Be careful about the directions of the currents in the components to maintain consistency. For loop 2, I2 going through the 2k resistor R2 should be a voltage drop, while I1 going in the opposite direction through the 2k would be a voltage rise. So

Loop 2: 10V - (I2 - I1)R2 - I2R3 - (I2 - I3)R4 - 6V = 0

Do the same for Loops 1 & 3.

 

[boo_lufc]

so does this look right:

Loop 1: 12 - I1R1 - (I1-I2)R2 - 10 = 0

Loop 3: 6 - (I3-I2)R4 - I3R5 = 0

with loop 2 given in your last post

 

[gneill]

That looks okay to me. Solve for the currents and let's see what you get.

 

[boo_lufc]

I have got:

I1 = 0.782mA

I2 = 0.955mA

I3 = 1.391mA Do these seem right?

When I use these values to work out the current across R2 for example I get the same value but negative and positive depending on what loop eqn I use. This seems like it would be right but in terms of the question: Find all the currents in the cct: how would I present this as a current in the cct?

 

[gneill]

I have got:

I1 = 0.782mA

I2 = 0.955mA

I3 = 1.391mA Do these seem right?

They aren't the values that I I get... For example, I get 2mA for I2.

When I use these values to work out the current across R2 for example I get the same value but negative and positive depending on what loop eqn I use. This seems like it would be right but in terms of the question: Find all the currents in the cct: how would I present this as a current in the cct?

I think you'll want to indicate branch currents that will show the current for every component. You might choose the current directions so that the values are all positive.

 

[boo_lufc]

From the equation for loop 3, i rearanged to get I2 = 5*I3 - 6 then subbed this value into the loop equations for lopp 1 and 2 and solved simulatneous equations.

Could you confirm if firstly this equation of I2 i have subbed in is correct and then if the method for finding the currents is correct please?

If this is right I must have just made an error in the algebra or calculations and can look again.

 

[gneill]

Rearranging equation 3 to solve for I2 and then substituting that value into the other equations is fine in theory. But without seeing how you've gone about doing it and the subsequent algebra, I can't spot an error in what I can't see.

You'll have to show your work.

 

[boo_lufc]

Ok if I take it step by step then should be able to spot mistakes.
First I just put the resistor values in and canceled the loop equations as shown:

Loop 1: 12 - I1*R1 - (I1 - I2)*R2 - 10 = 0
2 - 3I1 - 2I1 + 2I2 = 0
2 - 5I1 + 2I2 = 0

Loop 2: 10 - (I2 - I1)*R2 - I2*R3 - (I2 - I3)*R4 - 6 = 0
4 - 2I2 + 2I1 - I2 - I2 + I3 = 0
4 - 4I2 + 2I1 + I3 = 0

Loop 3: 6 - (I3 - I2)*R4 - I3*R5 = 0
6 - I3 + I2 - 4*I3 = 0
6 - 5I3 + I2 = 0
I2 = 5I3 - 6

 

[gneill]

That looks good so far.

 

[boo_lufc]

Ok so i then replace I2 with (5*I3 - 6) in the loop equations for loops 1 and 2:

Loop1: 2 - 5I1 + 2[5I3 - 6] = 0
2 - 5I1 + 10I3 - 12 = 0
-10 -5I1 + 10I3 = 0
10I3 - 5I1 = 10 -----------------(1)

Loop 2: 4 - 4[5I3 - 6] - 2I1 + I3 = 0
4 - 20I3 + 24 - 2I1 + I3 = 0
28 - 19I3 -2I1 = 0
28 = 19I3 - 2I1
19I3 - 2I1 = 28------------------(2)

 

[boo_lufc]

I think i have spotted my mistake, you say you had 2mA for I2, did you also get
I1 = 1.2mA and
I3 = 1.6mA?

 

[gneill]

That looks good!

 

[boo_lufc]

Ok good, so then to work out the voltage across all resistors using ohms law:
VR1 = I1*3 = 3.6V
VR2 = (I1 - I2)*2 = -1.6V
VR3 = I2*1 = 2V
VR4 = (I3 - I2)*R4 = -0.4V
VR5 = I3*4 = 6.4V

With VR2 and VR4 being negative or positive depending what loop you use